A system is know to be in the normalized state described by the wave function
=\frac{1}{\sqrt{30}}(5Y_4^3+Y_6^3-2Y_6^0) "\psi(\theta,\varphi)=\frac{1}{\sqrt{30}}(5Y_4^3+Y_6^3-2Y_6^0)")
Where "Y_l^m (\theta,\varphi)")
are the spherical harmonics. The probability of finding the system in a state with azimuthal orbital quantum number m = 3 is
A. 0
B. 1/15
C. 1/6
D. 1/3
E. 13/15
For the state}\right&space;\rangle "\left | \alpha\right \rangle = \sum_{i}c_i\left | a^{(i)}\right \rangle")
, probability is 
States with m = 3 →
and 
with c1 = 5/√30 and c2 = 1/√30
Therefore,
Answer: E
Where
A. 0
B. 1/15
C. 1/6
D. 1/3
E. 13/15
(GR9277 #28)
Solution:For the state
States with m = 3 →
Therefore,
Answer: E
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